2 3 X 7 10

$\exponential{(ten)}{ii} + 7 x + 10 $

\left(x+two\correct)\left(x+5\right)

\left(x+two\right)\left(ten+5\right)

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a+b=seven ab=i\times 10=ten

Cistron the expression by grouping. Commencement, the expression needs to be rewritten as x^{2}+ax+bx+10. To notice a and b, set upwards a organization to be solved.

1,10 ii,5

Since ab is positive, a and b take the same sign. Since a+b is positive, a and b are both positive. Listing all such integer pairs that requite product x.

1+10=11 ii+5=seven

Calculate the sum for each pair.

a=2 b=5

The solution is the pair that gives sum vii.

\left(x^{2}+2x\correct)+\left(5x+ten\right)

Rewrite 10^{2}+7x+ten as \left(x^{ii}+2x\right)+\left(5x+10\right).

ten\left(ten+2\right)+five\left(x+ii\right)

Gene out x in the starting time and 5 in the second group.

\left(x+two\correct)\left(10+5\right)

Factor out common term x+2 by using distributive property.

x^{2}+7x+ten=0

Quadratic polynomial can exist factored using the transformation ax^{two}+bx+c=a\left(x-x_{one}\right)\left(x-x_{2}\right), where x_{1} and x_{two} are the solutions of the quadratic equation ax^{ii}+bx+c=0.

ten=\frac{-7±\sqrt{7^{two}-4\times x}}{2}

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{two}-4ac}}{2a}. The quadratic formula gives ii solutions, one when ± is improver and i when it is subtraction.

10=\frac{-7±\sqrt{49-4\times x}}{2}

Square 7.

x=\frac{-7±\sqrt{49-40}}{2}

Multiply -4 times 10.

x=\frac{-7±\sqrt{9}}{2}

Add 49 to -40.

x=\frac{-7±3}{2}

Take the square root of ix.

x=\frac{-4}{2}

Now solve the equation x=\frac{-vii±3}{2} when ± is plus. Add -vii to 3.

x=\frac{-10}{2}

Now solve the equation 10=\frac{-7±3}{2} when ± is minus. Decrease three from -vii.

x^{2}+7x+x=\left(x-\left(-2\correct)\right)\left(x-\left(-5\right)\correct)

Factor the original expression using ax^{two}+bx+c=a\left(ten-x_{i}\right)\left(10-x_{2}\right). Substitute -2 for x_{i} and -5 for x_{ii}.

x^{2}+7x+x=\left(x+2\correct)\left(x+5\right)

Simplify all the expressions of the form p-\left(-q\right) to p+q.

ten ^ 2 +7x +10 = 0

Quadratic equations such equally this one can be solved by a new direct factoring method that does non require guess work. To employ the direct factoring method, the equation must be in the form 10^2+Bx+C=0.

r + s = -7 rs = 10

Let r and s be the factors for the quadratic equation such that x^ii+Bx+C=(x−r)(10−due south) where sum of factors (r+south)=−B and the product of factors rs = C

r = -\frac{seven}{2} - u s = -\frac{vii}{ii} + u

Two numbers r and s sum upward to -7 exactly when the average of the two numbers is \frac{one}{2}*-7 = -\frac{7}{2}. You tin also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented past the quadratic equation y=ten^2+Bx+C. The values of r and due south are equidistant from the center by an unknown quantity u. Express r and due south with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.internet/customsolver/quadraticgraph.png' mode='width: 100%;max-width: 700px' /></div>

(-\frac{seven}{2} - u) (-\frac{7}{2} + u) = 10

To solve for unknown quantity u, substitute these in the product equation rs = 10

\frac{49}{4} - u^2 = 10

Simplify past expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 10-\frac{49}{4} = -\frac{ix}{iv}

Simplify the expression by subtracting \frac{49}{iv} on both sides

u^2 = \frac{ix}{4} u = \pm\sqrt{\frac{nine}{4}} = \pm \frac{3}{ii}

Simplify the expression by multiplying -1 on both sides and take the foursquare root to obtain the value of unknown variable u

r =-\frac{vii}{two} - \frac{3}{2} = -v s = -\frac{7}{2} + \frac{3}{2} = -two

The factors r and southward are the solutions to the quadratic equation. Substitute the value of u to compute the r and south.

2 3 X 7 10,

Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20%2B7x%2B10

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